General Trends and Descriptive Analysis

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Abstract

This report explores the relationships between various factors influencing student performance, using exploratory data analysis (EDA) to identify key trends and correlations. The analysis focuses on variables such as study habits, access to resources, parental involvement, and environmental factors, and how they impact final exam scores. Insights gained from the data will inform recommendations aimed at improving academic outcomes for students.

Data Source

The dataset was sourced from Kaggle under the CC0 1.0 universal “No Copyright” license. We are free to copy, modify, distribute and perform the work, even for commercial purposes, all without asking permission. Learn more about this license here here.

URL for data in Kaggle: Student Performance Factors Dataset

Import and View Data

student_data <- read.csv('../data/StudentPerformanceFactors.csv', header = TRUE)
student_data    # Display the dataset

Summary Statistics

# Summary statistics for all variables
summary(student_data)   # Summary statistics for all variables

 Hours_Studied     Attendance     Parental_Involvement Access_to_Resources
 Min.   : 1.00   Min.   : 60.00   Length:6607          Length:6607        
 1st Qu.:16.00   1st Qu.: 70.00   Class :character     Class :character   
 Median :20.00   Median : 80.00   Mode  :character     Mode  :character   
 Mean   :19.98   Mean   : 79.98                                           
 3rd Qu.:24.00   3rd Qu.: 90.00                                           
 Max.   :44.00   Max.   :100.00                                           
 Extracurricular_Activities  Sleep_Hours     Previous_Scores 
 Length:6607                Min.   : 4.000   Min.   : 50.00  
 Class :character           1st Qu.: 6.000   1st Qu.: 63.00  
 Mode  :character           Median : 7.000   Median : 75.00  
                            Mean   : 7.029   Mean   : 75.07  
                            3rd Qu.: 8.000   3rd Qu.: 88.00  
                            Max.   :10.000   Max.   :100.00  
 Motivation_Level   Internet_Access    Tutoring_Sessions Family_Income     
 Length:6607        Length:6607        Min.   :0.000     Length:6607       
 Class :character   Class :character   1st Qu.:1.000     Class :character  
 Mode  :character   Mode  :character   Median :1.000     Mode  :character  
                                       Mean   :1.494                       
                                       3rd Qu.:2.000                       
                                       Max.   :8.000                       
 Teacher_Quality    School_Type        Peer_Influence     Physical_Activity
 Length:6607        Length:6607        Length:6607        Min.   :0.000    
 Class :character   Class :character   Class :character   1st Qu.:2.000    
 Mode  :character   Mode  :character   Mode  :character   Median :3.000    
                                                          Mean   :2.968    
                                                          3rd Qu.:4.000    
                                                          Max.   :6.000    
 Learning_Disabilities Parental_Education_Level Distance_from_Home
 Length:6607           Length:6607              Length:6607       
 Class :character      Class :character         Class :character  
 Mode  :character      Mode  :character         Mode  :character  
                                                                  
                                                                  
                                                                  
    Gender            Exam_Score    
 Length:6607        Min.   : 55.00  
 Class :character   1st Qu.: 65.00  
 Mode  :character   Median : 67.00  
                    Mean   : 67.24  
                    3rd Qu.: 69.00  
                    Max.   :101.00  

head(student_data)      # Display the first few rows of the dataset

str(student_data)       # Structure of the dataset

'data.frame':   6607 obs. of  20 variables:
 $ Hours_Studied             : int  23 19 24 29 19 19 29 25 17 23 ...
 $ Attendance                : int  84 64 98 89 92 88 84 78 94 98 ...
 $ Parental_Involvement      : chr  "Low" "Low" "Medium" "Low" ...
 $ Access_to_Resources       : chr  "High" "Medium" "Medium" "Medium" ...
 $ Extracurricular_Activities: chr  "No" "No" "Yes" "Yes" ...
 $ Sleep_Hours               : int  7 8 7 8 6 8 7 6 6 8 ...
 $ Previous_Scores           : int  73 59 91 98 65 89 68 50 80 71 ...
 $ Motivation_Level          : chr  "Low" "Low" "Medium" "Medium" ...
 $ Internet_Access           : chr  "Yes" "Yes" "Yes" "Yes" ...
 $ Tutoring_Sessions         : int  0 2 2 1 3 3 1 1 0 0 ...
 $ Family_Income             : chr  "Low" "Medium" "Medium" "Medium" ...
 $ Teacher_Quality           : chr  "Medium" "Medium" "Medium" "Medium" ...
 $ School_Type               : chr  "Public" "Public" "Public" "Public" ...
 $ Peer_Influence            : chr  "Positive" "Negative" "Neutral" "Negative" ...
 $ Physical_Activity         : int  3 4 4 4 4 3 2 2 1 5 ...
 $ Learning_Disabilities     : chr  "No" "No" "No" "No" ...
 $ Parental_Education_Level  : chr  "High School" "College" "Postgraduate" "High School" ...
 $ Distance_from_Home        : chr  "Near" "Moderate" "Near" "Moderate" ...
 $ Gender                    : chr  "Male" "Female" "Male" "Male" ...
 $ Exam_Score                : int  67 61 74 71 70 71 67 66 69 72 ...

sum(is.na(student_data))    # Check for missing values

[1] 0

Data Cleaning

We now check whether the dataset contains any missing values and remove them if necessary.

# Check missing values in each column
colSums(is.na(student_data))

             Hours_Studied                 Attendance 
                         0                          0 
      Parental_Involvement        Access_to_Resources 
                         0                          0 
Extracurricular_Activities                Sleep_Hours 
                         0                          0 
           Previous_Scores           Motivation_Level 
                         0                          0 
           Internet_Access          Tutoring_Sessions 
                         0                          0 
             Family_Income            Teacher_Quality 
                         0                          0 
               School_Type             Peer_Influence 
                         0                          0 
         Physical_Activity      Learning_Disabilities 
                         0                          0 
  Parental_Education_Level         Distance_from_Home 
                         0                          0 
                    Gender                 Exam_Score 
                         0                          0 

We can see that there are no missing values in the dataset.

We now check the unique values in each categorical column to identify any inconsistencies.

# Unique values in each categorical column
lapply(student_data[, sapply(student_data, is.character)], unique)

$Parental_Involvement
[1] "Low"    "Medium" "High"  

$Access_to_Resources
[1] "High"   "Medium" "Low"   

$Extracurricular_Activities
[1] "No"  "Yes"

$Motivation_Level
[1] "Low"    "Medium" "High"  

$Internet_Access
[1] "Yes" "No" 

$Family_Income
[1] "Low"    "Medium" "High"  

$Teacher_Quality
[1] "Medium" "High"   "Low"    ""      

$School_Type
[1] "Public"  "Private"

$Peer_Influence
[1] "Positive" "Negative" "Neutral" 

$Learning_Disabilities
[1] "No"  "Yes"

$Parental_Education_Level
[1] "High School"  "College"      "Postgraduate" ""            

$Distance_from_Home
[1] "Near"     "Moderate" "Far"      ""        

$Gender
[1] "Male"   "Female"

Low
Medium
High

High
Medium
Low

No
Yes

Low
Medium
High

Yes
No

Low
Medium
High

Medium
High
Low

Public
Private

Positive
Negative
Neutral

No
Yes

High School
College
Postgraduate

Near
Moderate
Far

Male
Female

From the results above we can Teacher_Quality, Parental_Education_Level and Distance_from_Home have missing values. We will now investigate further to see exactly what these missing values are.

# Check for missing values in Teacher_Quality
teacher_quality_missing <- student_data[student_data$Teacher_Quality == "",]
teacher_quality_missing

nrow(teacher_quality_missing)

[1] 78

We can see that only 78 rows have missing values in the Teacher_Quality column. We will now investigate the Parental_Education_Level column.

# Check for missing values in Parental_Education_Level
parental_education_level_missing <- student_data[student_data$Parental_Education_Level == "",]
parental_education_level_missing

nrow(parental_education_level_missing)

[1] 90

We can see that 90 rows have missing values in the Parental_Education_Level column. We will now investigate the Distance_from_Home column.

# Check for missing values in Distance_from_Home
distance_from_home_missing <- student_data[student_data$Distance_from_Home == "",]
distance_from_home_missing

nrow(distance_from_home_missing)

[1] 67

We can see that 67 rows have missing values in the Distance_from_Home column. All the missing values combined make up less than 10% of the dataset. We will remove these rows from the dataset.

# Remove rows with missing values
student_data <- subset(student_data, Teacher_Quality != "" & Parental_Education_Level != "" & Distance_from_Home != "")

# Check for missing values
lapply(student_data[, sapply(student_data, is.character)], unique)

$Parental_Involvement
[1] "Low"    "Medium" "High"  

$Access_to_Resources
[1] "High"   "Medium" "Low"   

$Extracurricular_Activities
[1] "No"  "Yes"

$Motivation_Level
[1] "Low"    "Medium" "High"  

$Internet_Access
[1] "Yes" "No" 

$Family_Income
[1] "Low"    "Medium" "High"  

$Teacher_Quality
[1] "Medium" "High"   "Low"   

$School_Type
[1] "Public"  "Private"

$Peer_Influence
[1] "Positive" "Negative" "Neutral" 

$Learning_Disabilities
[1] "No"  "Yes"

$Parental_Education_Level
[1] "High School"  "College"      "Postgraduate"

$Distance_from_Home
[1] "Near"     "Moderate" "Far"     

$Gender
[1] "Male"   "Female"

Low
Medium
High

High
Medium
Low

No
Yes

Low
Medium
High

Yes
No

Low
Medium
High

Medium
High
Low

Public
Private

Positive
Negative
Neutral

No
Yes

High School
College
Postgraduate

Near
Moderate
Far

Male
Female

We now investigate the dependent variable, Final_Exam_Score, to identify any outliers.

# Outliers in Final_Exam_Score
outliers_in_exam_score <- student_data[student_data$Exam_Score > 100,]
outliers_in_exam_score

nrow(outliers_in_exam_score)

[1] 1

We can see that only 1 student got exam score of 101 which is an outlier. We will remove this row from the dataset.

# Remove outliers
student_data <- subset(student_data, Exam_Score <= 100)

Now that the data has been cleaned, we can proceed with the exploratory data analysis.

Distribution of Final Exam Scores

Distribution of Final Exam Scores (Without Considering Other Factors)

Here we will explore the distribution of final exam scores among students with without considering other factors. To find our the distribution of final exam scores, we first need to sample the data and plot a histogram.

# Sample the data
set.seed(123)
exam_score_sample <- student_data$Exam_Score[sample(nrow(student_data), 100)]
exam_score_sample

  [1] 66 70 72 68 62 69 72 66 58 69 63 67 70 64 65 70 75 70 69 69 68 63 75 69 68
 [26] 66 68 70 69 64 60 67 66 70 69 65 65 69 69 66 64 64 66 66 66 72 61 71 66 65
 [51] 63 69 70 73 66 70 64 68 71 69 63 68 63 65 70 66 71 71 87 72 67 66 71 64 67
 [76] 63 72 64 68 66 75 70 64 67 65 66 63 69 68 65 68 65 61 71 69 68 66 61 59 65

# Plot histogram
hist(exam_score_sample, main = "Distribution of Final Exam Scores", xlab = "Final Exam Score", col = "skyblue", border = "black")

From the histogram, we can see that the distribution of final exam scores is approximately normal. We now plot a boxplot to visualize the spread of scores and identify any outliers.

# Boxplot of final exam scores
boxplot(exam_score_sample, main = "Boxplot of Final Exam Scores", col = "skyblue", border = "black")

The boxplot shows that the distribution of final exam scores is centered around the median, with a few outliers on the lower end of the scale. Now we use numerical methods to confirm the normality of the distribution.

# Shapiro-Wilk test for normality
shapiro.test(exam_score_sample)

    Shapiro-Wilk normality test

data:  exam_score_sample
W = 0.92839, p-value = 4.035e-05

The Shapiro-Wilk test confirms that the distribution of final exam scores is not normal, with a p-value less than 0.05.

Distribution of Final Exam Scores (Parental_Involvement Category)

Next, we explore the distribution of final exam scores based on parental involvement levels. We will create a boxplot to compare the scores of students with different levels of parental involvement.

# Sample the data
high_parental_involvement <- student_data$Exam_Score[student_data$Parental_Involvement == "High"][sample(sum(student_data$Parental_Involvement == "High"), 100)]
medium_parental_involvement <- student_data$Exam_Score[student_data$Parental_Involvement == "Medium"][sample(sum(student_data$Parental_Involvement == "Medium"), 100)]
low_parental_involvement <- student_data$Exam_Score[student_data$Parental_Involvement == "Low"][sample(sum(student_data$Parental_Involvement == "Low"), 100)]

# Histogram of final exam scores by parental involvement
par(mfrow = c(1, 3))
hist(high_parental_involvement, main = "High Parental Involvement", xlab = "Final Exam Score", col = "skyblue", border = "black")
hist(medium_parental_involvement, main = "Medium Parental Involvement", xlab = "Final Exam Score", col = "skyblue", border = "black")
hist(low_parental_involvement, main = "Low Parental Involvement", xlab = "Final Exam Score", col = "skyblue", border = "black")

The three histograms show the distribution of final exam scores for students with high, medium, and low levels of parental involvement. We can see that the distribution of the scores seems to be similar across all three categories. They seem to follow a normal distribution, with a slight skew towards higher scores for students with high parental involvement. We now use numerical methods to confirm the normality of the distributions.

# Shapiro-Wilk test for normality
shapiro.test(high_parental_involvement)

    Shapiro-Wilk normality test

data:  high_parental_involvement
W = 0.97588, p-value = 0.06325

shapiro.test(medium_parental_involvement)

    Shapiro-Wilk normality test

data:  medium_parental_involvement
W = 0.97777, p-value = 0.08891

shapiro.test(low_parental_involvement)

    Shapiro-Wilk normality test

data:  low_parental_involvement
W = 0.98902, p-value = 0.5864

The Shapiro-Wilk test confirms that the distributions of final exam scores for students with high and medium levels of parental involvement are approximately normal, with p-values greater than 0.05. However, the distribution for students with low parental involvement is slightly skewed, with a p-value less than 0.05.

We now investigate the distribution of final exam scores for students with low parental involvement further. We will now plot a density plot to visualize the distribution more clearly.

# Density plot of final exam scores for students with low parental involvement
plot(density(low_parental_involvement), main = "Density Plot of Final Exam Scores for Low Parental Involvement", xlab = "Final Exam Score", col = "skyblue")

The density plot shows that the distribution of final exam is slightly skewed to the left for students with low parental involvement. We will now create a QQ plot to compare the distribution of scores to a normal distribution.

# QQ plot of final exam scores for students with low parental involvement
qqnorm(low_parental_involvement, main = "QQ Plot of Final Exam Scores for Low Parental Involvement", col = "skyblue")
qqline(low_parental_involvement, col = "red")

The QQ plot confirms that the distribution of final exam scores for students with low parental involvement is slightly skewed to the left, deviating from a normal distribution.

Distribution of Final Exam Scores (Access_to_Resources Category)

Next, we explore the distribution of final exam scores based on access to resources. We will create a boxplot to compare the scores of students with different levels of access to resources.

# Sample the data
high_access_to_resources <- student_data$Exam_Score[student_data$Access_to_Resources == "High"][sample(sum(student_data$Access_to_Resources == "High"), 100)]
medium_access_to_resources <- student_data$Exam_Score[student_data$Access_to_Resources == "Medium"][sample(sum(student_data$Access_to_Resources == "Medium"), 100)]
low_access_to_resources <- student_data$Exam_Score[student_data$Access_to_Resources == "Low"][sample(sum(student_data$Access_to_Resources == "Low"), 100)]

# Histogram of final exam scores by access to resources
par(mfrow = c(1, 3))
hist(high_access_to_resources, main = "High Access to Resources", xlab = "Final Exam Score", col = "skyblue", border = "black")
hist(medium_access_to_resources, main = "Medium Access to Resources", xlab = "Final Exam Score", col = "skyblue", border = "black")
hist(low_access_to_resources, main = "Low Access to Resources", xlab = "Final Exam Score", col = "skyblue", border = "black")

plot(density(low_access_to_resources), main = "Low Access to Resources", xlab = "Final Exam Score", col = "skyblue")
plot(density(medium_access_to_resources), main = "Medium Access to Resources", xlab = "Final Exam Score", col = "skyblue")
plot(density(high_access_to_resources), main = "High Access to Resources", xlab = "Final Exam Score", col = "skyblue")

The histograms and density plots show the distribution of final exam scores for students with high, medium, and low levels of access to resources. The distributions seem to be similar across all three categories, with a slight skew towards higher scores for students with high access to resources. We will now use numerical methods to confirm the normality of the distributions.

# Shapiro-Wilk test for normality
shapiro.test(high_access_to_resources)

    Shapiro-Wilk normality test

data:  high_access_to_resources
W = 0.97748, p-value = 0.0844

shapiro.test(medium_access_to_resources)

    Shapiro-Wilk normality test

data:  medium_access_to_resources
W = 0.97329, p-value = 0.03968

shapiro.test(low_access_to_resources)

    Shapiro-Wilk normality test

data:  low_access_to_resources
W = 0.97248, p-value = 0.0343

The Shapiro-Wilk test shows that the distribution of all three categories of access to resources is not normal, with p-values less than 0.05. This indicates that the distribution of final exam scores is skewed for students with different levels of access to resources.

Distribution of Final Exam Scores (Extracurricular_Activities Category)

Next, we explore the distribution of final exam scores based on participation in extracurricular activities. We will create a boxplot to compare the scores of students who participate in extracurricular activities and those who do not.

# Sample the data
participate_extracurricular <- student_data$Exam_Score[student_data$Extracurricular_Activities == "Yes"][sample(sum(student_data$Extracurricular_Activities == "Yes"), 100)]
do_not_participate_extracurricular <- student_data$Exam_Score[student_data$Extracurricular_Activities == "No"][sample(sum(student_data$Extracurricular_Activities == "No"), 100)]

# Boxplot of final exam scores by extracurricular activities
boxplot(student_data$Exam_Score ~ student_data$Extracurricular_Activities, main = "Final Exam Scores by Extracurricular Activities", xlab = "Extracurricular Activities", ylab = "Final Exam Score", col = "skyblue", border = "black")

The boxplot shows that students who participate in extracurricular activities tend to have higher final exam scores compared to those who do not. Now we will visualize the distribution of scores for both groups using histograms.

# Histogram of final exam scores by extracurricular activities
par(mfrow = c(1, 2))
hist(participate_extracurricular, main = "Extracurricular Activities", xlab = "Final Exam Score", col = "skyblue", border = "black")
hist(do_not_participate_extracurricular, main = "No Extracurricular Activities", xlab = "Final Exam Score", col = "skyblue", border = "black")

Both histograms show right skewed distributions, with students who participate in extracurricular activities having higher final exam scores. We will now use numerical methods to confirm the normality of the distributions with the following hypothesis test.

• Null Hypothesis (\(H_0\)): The distribution of final exam scores for students who participate in extracurricular activities is normal.
• Alternative Hypothesis (\(H_1\)): The distribution of final exam scores for students who participate in extracurricular activities is not normal.

# Shapiro-Wilk test for normality
shapiro.test(participate_extracurricular)

    Shapiro-Wilk normality test

data:  participate_extracurricular
W = 0.97923, p-value = 0.1158

The Shapiro-Wilk test confirms that the distribution of final exam scores for students who participate in extracurricular activities is not normal, with a p-value less than 0.05. Thus, we reject the null hypothesis.

Distribution of Final Exam Scores (Motivation_Level Category)

Next we explore the distribution of final exam scores based on motivation levels. We will create a histogram to compare the scores of students with different motivation levels.

# Smaple Data
high_motivation <- student_data$Exam_Score[student_data$Motivation_Level == "High"][sample(sum(student_data$Motivation_Level == "High"), 100)]
medium_motivation <- student_data$Exam_Score[student_data$Motivation_Level == "High"][sample(sum(student_data$Motivation_Level == "Medium"), 100)]
low_motivation <- student_data$Exam_Score[student_data$Motivation_Level == "High"][sample(sum(student_data$Motivation_Level == "Low"), 100)]

# Histogram of final exam scores by motivation level
par(mfrow = c(1, 3))
hist(high_motivation, main = "High Motivation Level", xlab = "Final Exam Score", col = "skyblue", border = "black")
hist(medium_motivation, main = "Medium Motivation Level", xlab = "Final Exam Score", col = "skyblue", border = "black")
hist(low_motivation, main = "Low Motivation Level", xlab = "Final Exam Score", col = "skyblue", border = "black")

The histograms for High and medium motivation levels show a normal distribution of final exam scores, while the low motivation level histogram shows a right-skewed distribution. We will now use numerical methods to confirm the normality of the distributions.

# Shapiro-Wilk test for normality
shapiro.test(high_motivation)

    Shapiro-Wilk normality test

data:  high_motivation
W = 0.94412, p-value = 0.0003468

shapiro.test(medium_motivation)

    Shapiro-Wilk normality test

data:  medium_motivation
W = 0.94732, p-value = 0.07268

shapiro.test(low_motivation)

    Shapiro-Wilk normality test

data:  low_motivation
W = 0.83108, p-value = 5.376e-07

The Shapiro-Wilk test confirms that the distributions of final exam scores for students with high and medium motivation levels are approximately normal, with p-values greater than 0.05. However, the distribution for students with low motivation levels is slightly skewed, with a p-value less than 0.05.

Distribution of Final Exam Scores (Internet_Access Category)

Next, we explore the distribution of final exam scores based on internet access. We will create a boxplot to compare the scores of students with and without internet access.

# Sample the data
internet_access <- student_data$Exam_Score[student_data$Internet_Access == "Yes"][sample(sum(student_data$Internet_Access == "Yes"), 100)]
no_internet_access <- student_data$Exam_Score[student_data$Internet_Access == "No"][sample(sum(student_data$Internet_Access == "No"), 100)]

# Boxplot of final exam scores by internet access
boxplot(student_data$Exam_Score ~ student_data$Internet_Access, main = "Final Exam Scores by Internet Access", xlab = "Internet Access", ylab = "Final Exam Score", col = "skyblue", border = "black")

# Histogram of final exam scores by internet access
par(mfrow = c(1, 2))
hist(internet_access, main = "Internet Access", xlab = "Final Exam Score", col = "skyblue", border = "black")
hist(no_internet_access, main = "No Internet Access", xlab = "Final Exam Score", col = "skyblue", border = "black")

# Shapiro-Wilk test for normality
shapiro.test(internet_access)

    Shapiro-Wilk normality test

data:  internet_access
W = 0.96856, p-value = 0.01718

shapiro.test(no_internet_access)

    Shapiro-Wilk normality test

data:  no_internet_access
W = 0.93958, p-value = 0.0001819

The boxplot shows that students with internet access tend to have higher final exam scores compared to those without internet access. The histograms show that the distribution of final exam scores for students with no internet access is approximately normal, while the distribution for students with internet access is slightly skewed. The Shapiro-Wilk test confirms that the distribution of final exam scores for students without internet access is normal, with a p-value greater than 0.05. However, the distribution for students with internet access is slightly skewed, with a p-value less than 0.05.

Summary of Distribution Analysis

In summary, the distribution of final exam scores is approximately normal when considering all students. However, when examining the scores based on other factors such as parental involvement, access to resources, extracurricular activities, and motivation levels, the distributions vary. Students with high parental involvement and high access to resources tend to have higher final exam scores, while students who participate in extracurricular activities also perform better. Motivation levels also play a role in student performance, with students who are highly motivated achieving higher scores.

Average number of hours students study per week

Average Number of Hours Students Study per Week (Without Considering Other Factors)

Next, we explore the average number of hours students study per week without considering other factors.

# Summary statistics for study hours
summary(student_data$Hours_Studied)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   1.00   16.00   20.00   19.98   24.00   44.00 

The summary statistics show that the average number of hours students study per week is approximately 19.98 hours. We will now visualize the distribution of study hours using a histogram.

# Histogram of study hours
hist(student_data$Hours_Studied, main = "Distribution of Study Hours", xlab = "Study Hours", col = "skyblue", border = "black")

The distribution of study hours is approximately normal. We will now use numerical methods to confirm the normality of the distribution.

# Shapiro-Wilk test for normality
sample_hours_studied <- student_data$Hours_Studied[sample(nrow(student_data), 100)]
shapiro.test(sample_hours_studied)

    Shapiro-Wilk normality test

data:  sample_hours_studied
W = 0.99057, p-value = 0.7103

The p-value is greater than 0.05, indicating that the distribution of study hours is approximately normal.

Average Number of Hours Students Study per Week (Parental_Involvement Category)

Next, we will explore the average number of hours students study per week based on parental involvement levels.

# Summary statistics for study hours by parental involvement
summary(student_data$Hours_Studied[student_data$Parental_Involvement == "High"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   1.00   16.00   20.00   19.86   24.00   44.00 

summary(student_data$Hours_Studied[student_data$Parental_Involvement == "Medium"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   1.00   16.00   20.00   19.99   24.00   39.00 

summary(student_data$Hours_Studied[student_data$Parental_Involvement == "Low"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   2.00   16.00   20.00   20.11   24.00   38.00 

The summary statistics of the average number of hours students study per week based on parental involvement levels do not show significant differences. We will now use ANOVA to test for differences in study hours based on parental involvement levels. But first, we need to check the assumptions of ANOVA.

# Sample the data
high_parental_involvement <- student_data$Hours_Studied[student_data$Parental_Involvement == "High"][sample(sum(student_data$Parental_Involvement == "High"), 100)]
medium_parental_involvement <- student_data$Hours_Studied[student_data$Parental_Involvement == "Medium"][sample(sum(student_data$Parental_Involvement == "Medium"), 100)]
low_parental_involvement <- student_data$Hours_Studied[student_data$Parental_Involvement == "Low"][sample(sum(student_data$Parental_Involvement == "Low"), 100)]

# Histogram of study hours by parental involvement
par(mfrow = c(1, 3))
hist(high_parental_involvement, main = "High Parental Involvement", xlab = "Study Hours", col = "skyblue", border = "black")
hist(medium_parental_involvement, main = "Medium Parental Involvement", xlab = "Study Hours", col = "skyblue", border = "black")
hist(low_parental_involvement, main = "Low Parental Involvement", xlab = "Study Hours", col = "skyblue", border = "black")

# Check assumptions of ANOVA
shapiro.test(high_parental_involvement)

    Shapiro-Wilk normality test

data:  high_parental_involvement
W = 0.9827, p-value = 0.2145

shapiro.test(medium_parental_involvement)

    Shapiro-Wilk normality test

data:  medium_parental_involvement
W = 0.99085, p-value = 0.7331

shapiro.test(low_parental_involvement)

    Shapiro-Wilk normality test

data:  low_parental_involvement
W = 0.986, p-value = 0.3737

Both the histograms and the Shapiro-Wilk test show that the distribution of study hours is approximately normal for all students. We will now use ANOVA to test for differences in study hours based on parental involvement levels.

# ANOVA test for study hours by parental involvement
anova_parental_involvement <- aov(Hours_Studied ~ Parental_Involvement, data = student_data)
summary(anova_parental_involvement)

                       Df Sum Sq Mean Sq F value Pr(>F)
Parental_Involvement    2     49   24.44   0.682  0.506
Residuals            6374 228362   35.83               

The ANOVA test shows that there is no significant difference in the average number of hours students study per week based on parental involvement levels, with a p-value greater than 0.05. This indicates that parental involvement does not have a significant impact on the number of hours students study per week.

Average Number of Hours Students Study per Week (Access_to_Resources Category)

We now explore the average number of hours students study per week based on access to resources.

summary(student_data$Hours_Studied[student_data$Access_to_Resources == "High"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   1.00   16.00   20.00   20.02   24.00   39.00 

summary(student_data$Hours_Studied[student_data$Access_to_Resources == "Medium"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
    1.0    16.0    20.0    19.9    24.0    43.0 

summary(student_data$Hours_Studied[student_data$Access_to_Resources == "Low"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   2.00   16.00   20.00   20.12   24.00   44.00 

The summary statistics of the average number of hours students study per week based on access to resources do not show significant differences. We will now use ANOVA to test for differences in study hours based on access to resources. But first, we need to check the assumptions of ANOVA.

# Sample the data
high_access_to_resources <- student_data$Hours_Studied[student_data$Access_to_Resources == "High"][sample(sum(student_data$Access_to_Resources == "High"), 100)]
medium_access_to_resources <- student_data$Hours_Studied[student_data$Access_to_Resources == "Medium"][sample(sum(student_data$Access_to_Resources == "Medium"), 100)]
low_access_to_resources <- student_data$Hours_Studied[student_data$Access_to_Resources == "Low"][sample(sum(student_data$Access_to_Resources == "Low"), 100)]

# Histogram of study hours by access to resources
par(mfrow = c(1, 3))
hist(high_access_to_resources, main = "High Access to Resources", xlab = "Study Hours", col = "skyblue", border = "black")
hist(medium_access_to_resources, main = "Medium Access to Resources", xlab = "Study Hours", col = "skyblue", border = "black")
hist(low_access_to_resources, main = "Low Access to Resources", xlab = "Study Hours", col = "skyblue", border = "black")

# Check assumptions of ANOVA
shapiro.test(high_access_to_resources)

    Shapiro-Wilk normality test

data:  high_access_to_resources
W = 0.98025, p-value = 0.139

shapiro.test(medium_access_to_resources)

    Shapiro-Wilk normality test

data:  medium_access_to_resources
W = 0.96582, p-value = 0.01068

shapiro.test(low_access_to_resources)

    Shapiro-Wilk normality test

data:  low_access_to_resources
W = 0.97123, p-value = 0.02747

Both the histograms and the Shapiro-Wilk test show that the distribution of study hours is approximately normal for all students. We will now use ANOVA to test for differences in study hours based on access to resources.

# ANOVA test for study hours by access to resources
anova_access_to_resources <- aov(Hours_Studied ~ Access_to_Resources, data = student_data)
summary(anova_access_to_resources)

                      Df Sum Sq Mean Sq F value Pr(>F)
Access_to_Resources    2     48   24.12   0.673   0.51
Residuals           6374 228363   35.83               

The ANOVA test shows that there is no significant difference in the average number of hours students study per week based on access to resources, with a p-value greater than 0.05. This indicates that access to resources does not have a significant impact on the number of hours students study per week.

Average Number of Hours Students Study per Week (Extracurricular_Activities Category)

Next, we explore the average number of hours students study per week based on participation in extracurricular activities.

# Summary statistics for study hours by extracurricular activities
summary(student_data$Hours_Studied[student_data$Extracurricular_Activities == "Yes"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   1.00   16.00   20.00   19.93   24.00   43.00 

summary(student_data$Hours_Studied[student_data$Extracurricular_Activities == "No"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   2.00   16.00   20.00   20.04   24.00   44.00 

The summary statistics of the average number of hours students study per week based on participation in extracurricular activities show no significant differences. We will now use ANOVA to test for differences in study hours based on participation in extracurricular activities. But first, we need to check the assumptions of ANOVA.

# Sample the data
participate_extracurricular <- student_data$Hours_Studied[student_data$Extracurricular_Activities == "Yes"][sample(sum(student_data$Extracurricular_Activities == "Yes"), 100)]
do_not_participate_extracurricular <- student_data$Hours_Studied[student_data$Extracurricular_Activities == "No"][sample(sum(student_data$Extracurricular_Activities == "No"), 100)]

# Histogram of study hours by extracurricular activities
par(mfrow = c(1, 2))
hist(participate_extracurricular, main = "Extracurricular Activities", xlab = "Study Hours", col = "skyblue", border = "black")
hist(do_not_participate_extracurricular, main = "No Extracurricular Activities", xlab = "Study Hours", col = "skyblue", border = "black")

# Check assumptions of ANOVA
shapiro.test(participate_extracurricular)

    Shapiro-Wilk normality test

data:  participate_extracurricular
W = 0.98586, p-value = 0.3655

shapiro.test(do_not_participate_extracurricular)

    Shapiro-Wilk normality test

data:  do_not_participate_extracurricular
W = 0.98909, p-value = 0.5916

Both the histograms and the Shapiro-Wilk test show that the distribution of study hours is approximately normal for all students. We will now use ANOVA to test for differences in study hours based on participation in extracurricular activities.

# ANOVA test for study hours by extracurricular activities
anova_extracurricular_activities <- aov(Hours_Studied ~ Extracurricular_Activities, data = student_data)
summary(anova_extracurricular_activities)

                             Df Sum Sq Mean Sq F value Pr(>F)
Extracurricular_Activities    1     17   16.62   0.464  0.496
Residuals                  6375 228395   35.83               

The ANOVA test shows that there is no significant difference in the average number of hours students study per week based on participation in extracurricular activities, with a p-value greater than 0.05. This indicates that participation in extracurricular activities does not have a significant impact on the number of hours students study per week.

Average Number of Hours Students Study per Week (Motivation_Level Category)

Next, we explore the average number of hours students study per week based on motivation levels.

# Summary statistics for study hours by motivation level
summary(student_data$Hours_Studied[student_data$Motivation_Level == "High"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   1.00   16.00   20.00   19.75   24.00   39.00 

summary(student_data$Hours_Studied[student_data$Motivation_Level == "Medium"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   1.00   16.00   20.00   20.06   24.00   43.00 

summary(student_data$Hours_Studied[student_data$Motivation_Level == "Low"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   2.00   16.00   20.00   19.98   24.00   44.00 

The summary statistics of the average number of hours students study per week based on motivation levels show no significant differences. We will now use ANOVA to test for differences in study hours based on motivation levels. But first, we need to check the assumptions of ANOVA.

# Sample the data
high_motivation <- student_data$Hours_Studied[student_data$Motivation_Level == "High"][sample(sum(student_data$Motivation_Level == "High"), 100)]
medium_motivation <- student_data$Hours_Studied[student_data$Motivation_Level == "Medium"][sample(sum(student_data$Motivation_Level == "Medium"), 100)]
low_motivation <- student_data$Hours_Studied[student_data$Motivation_Level == "Low"][sample(sum(student_data$Motivation_Level == "Low"), 100)]

# Histogram of study hours by motivation level
par(mfrow = c(1, 3))
hist(high_motivation, main = "High Motivation Level", xlab = "Study Hours", col = "skyblue", border = "black")
hist(medium_motivation, main = "Medium Motivation Level", xlab = "Study Hours", col = "skyblue", border = "black")
hist(low_motivation, main = "Low Motivation Level", xlab = "Study Hours", col = "skyblue", border = "black")

# Check assumptions of ANOVA
shapiro.test(high_motivation)

    Shapiro-Wilk normality test

data:  high_motivation
W = 0.98676, p-value = 0.4217

shapiro.test(medium_motivation)

    Shapiro-Wilk normality test

data:  medium_motivation
W = 0.9649, p-value = 0.009127

shapiro.test(low_motivation)

    Shapiro-Wilk normality test

data:  low_motivation
W = 0.99065, p-value = 0.717

All three histograms and the Shapiro-Wilk test show that the distribution of study hours is approximately normal for all students. We will now use ANOVA to test for differences in study hours based on motivation levels.

# ANOVA test for study hours by motivation level
anova_motivation_level <- aov(Hours_Studied ~ Motivation_Level, data = student_data)
summary(anova_motivation_level)

                   Df Sum Sq Mean Sq F value Pr(>F)
Motivation_Level    2     90   45.14    1.26  0.284
Residuals        6374 228321   35.82               

The ANOVA test shows that there is no significant difference in the average number of hours students study per week based on motivation levels, with a p-value greater than 0.05. This indicates that motivation levels do not have a significant impact on the number of hours students study per week.

Average Number of Hours Students Study per Week (Internet Access Category)

Next, we explore the average number of hours students study per week based on internet access.

# Summary statistics for study hours by internet access
summary(student_data$Hours_Studied[student_data$Internet_Access == "Yes"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   1.00   16.00   20.00   19.99   24.00   44.00 

summary(student_data$Hours_Studied[student_data$Internet_Access == "No"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   4.00   15.00   20.00   19.83   24.00   37.00 

The summary statistics of the average number of hours students study per week based on internet access show no significant differences. We will now use ANOVA to test for differences in study hours based on internet access. But first, we need to check the assumptions of ANOVA.

# Sample the data
internet_access <- student_data$Hours_Studied[student_data$Internet_Access == "Yes"][sample(sum(student_data$Internet_Access == "Yes"), 100)]
no_internet_access <- student_data$Hours_Studied[student_data$Internet_Access == "No"][sample(sum(student_data$Internet_Access == "No"), 100)]

# Histogram of study hours by internet access
par(mfrow = c(1, 2))
hist(internet_access, main = "Internet Access", xlab = "Study Hours", col = "skyblue", border = "black")
hist(no_internet_access, main = "No Internet Access", xlab = "Study Hours", col = "skyblue", border = "black")

# Check assumptions of ANOVA
shapiro.test(internet_access)

    Shapiro-Wilk normality test

data:  internet_access
W = 0.99238, p-value = 0.8479

shapiro.test(no_internet_access)

    Shapiro-Wilk normality test

data:  no_internet_access
W = 0.99029, p-value = 0.6878

Both the histograms and the Shapiro-Wilk test show that the distribution of study hours is approximately normal for all students. We will now use ANOVA to test for differences in study hours based on internet access.

# ANOVA test for study hours by internet access
anova_internet_access <- aov(Hours_Studied ~ Internet_Access, data = student_data)
summary(anova_internet_access)

                  Df Sum Sq Mean Sq F value Pr(>F)
Internet_Access    1     11   11.39   0.318  0.573
Residuals       6375 228400   35.83               

The ANOVA test shows that there is no significant difference in the average number of hours students study per week based on internet access, with a p-value greater than 0.05. This indicates that internet access does not have a significant impact on the number of hours students study per week.

Average Number of Hours Students Study per Week (Family Income Category)

Next, we explore the average number of hours students study per week based on family income levels.

# Summary statistics for study hours by family income
summary(student_data$Hours_Studied[student_data$Family_Income == "Low"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   1.00   16.00   20.00   19.93   24.00   39.00 

summary(student_data$Hours_Studied[student_data$Family_Income == "Medium"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   1.00   16.00   20.00   20.07   24.00   44.00 

summary(student_data$Hours_Studied[student_data$Family_Income == "High"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   3.00   16.00   20.00   19.89   24.00   39.00 

The summary statistics of the average number of hours students study per week based on family income levels show no significant differences. We will now use ANOVA to test for differences in study hours based on family income levels. But first, we need to check the assumptions of ANOVA.

# Sample the data
low_family_income <- student_data$Hours_Studied[student_data$Family_Income == "Low"][sample(sum(student_data$Family_Income == "Low"), 100)]
medium_family_income <- student_data$Hours_Studied[student_data$Family_Income == "Medium"][sample(sum(student_data$Family_Income == "Medium"), 100)]
high_family_income <- student_data$Hours_Studied[student_data$Family_Income == "High"][sample(sum(student_data$Family_Income == "High"), 100)]

# Histogram of study hours by family income
par(mfrow = c(1, 3))
hist(low_family_income, main = "Low Family Income", xlab = "Study Hours", col = "skyblue", border = "black")
hist(medium_family_income, main = "Medium Family Income", xlab = "Study Hours", col = "skyblue", border = "black")
hist(high_family_income, main = "High Family Income", xlab = "Study Hours", col = "skyblue", border = "black")

# Check assumptions of ANOVA
shapiro.test(low_family_income)

    Shapiro-Wilk normality test

data:  low_family_income
W = 0.99118, p-value = 0.7595

shapiro.test(medium_family_income)

    Shapiro-Wilk normality test

data:  medium_family_income
W = 0.98624, p-value = 0.3883

shapiro.test(high_family_income)

    Shapiro-Wilk normality test

data:  high_family_income
W = 0.98195, p-value = 0.1881

The Shapiro-Wilk test shows that the distribution of study hours is approximately normal for students with medium and high family income levels, with p-values greater than 0.05. However, the distribution for students with low family income is slightly skewed, with a p-value less than 0.05 indicating that the distribution is not normal. We will now use Kruksal-Wallis test to test for differences in study hours based on family income levels.

# Kruskal-Wallis test for study hours by family income
kruskal.test(Hours_Studied ~ Family_Income, data = student_data)

    Kruskal-Wallis rank sum test

data:  Hours_Studied by Family_Income
Kruskal-Wallis chi-squared = 1.0581, df = 2, p-value = 0.5892

The Kruskal-Wallis test shows that there is no significant difference in the average number of hours students study per week based on family income levels, with a p-value greater than 0.05. This indicates that family income does not have a significant impact on the number of hours students study per week.

Average Number of Hours Students Study per Week (Teacher Quality Category)

Next, we explore the average number of hours students study per week based on teacher quality levels.

# Summary statistics for study hours by teacher quality
summary(student_data$Hours_Studied[student_data$Teacher_Quality == "Low"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   4.00   16.00   20.00   20.05   24.00   39.00 

summary(student_data$Hours_Studied[student_data$Teacher_Quality == "Medium"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   1.00   16.00   20.00   19.99   24.00   39.00 

summary(student_data$Hours_Studied[student_data$Teacher_Quality == "High"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   2.00   16.00   20.00   19.92   24.00   44.00 

The summary statistics of the average number of hours students study per week based on teacher quality levels show no significant differences. We will now use ANOVA to test for differences in study hours based on teacher quality levels. But first, we need to check the assumptions of ANOVA.

# Sample the data
low_teacher_quality <- student_data$Hours_Studied[student_data$Teacher_Quality == "Low"][sample(sum(student_data$Teacher_Quality == "Low"), 100)]
medium_teacher_quality <- student_data$Hours_Studied[student_data$Teacher_Quality == "Medium"][sample(sum(student_data$Teacher_Quality == "Medium"), 100)]
high_teacher_quality <- student_data$Hours_Studied[student_data$Teacher_Quality == "High"][sample(sum(student_data$Teacher_Quality == "High"), 100)]

# Histogram of study hours by teacher quality
par(mfrow = c(1, 3))
hist(low_teacher_quality, main = "Low Teacher Quality", xlab = "Study Hours", col = "skyblue", border = "black")
hist(medium_teacher_quality, main = "Medium Teacher Quality", xlab = "Study Hours", col = "skyblue", border = "black")
hist(high_teacher_quality, main = "High Teacher Quality", xlab = "Study Hours", col = "skyblue", border = "black")

# Check assumptions of ANOVA
shapiro.test(low_teacher_quality)

    Shapiro-Wilk normality test

data:  low_teacher_quality
W = 0.98158, p-value = 0.1763

shapiro.test(medium_teacher_quality)

    Shapiro-Wilk normality test

data:  medium_teacher_quality
W = 0.98216, p-value = 0.1952

shapiro.test(high_teacher_quality)

    Shapiro-Wilk normality test

data:  high_teacher_quality
W = 0.98864, p-value = 0.5568

All three histograms and the Shapiro-Wilk test show that the distribution of study hours is approximately normal for all students. We will now use ANOVA to test for differences in study hours based on teacher quality levels.

# ANOVA test for study hours by teacher quality
anova_teacher_quality <- aov(Hours_Studied ~ Teacher_Quality, data = student_data)
summary(anova_teacher_quality)

                  Df Sum Sq Mean Sq F value Pr(>F)
Teacher_Quality    2     12    5.88   0.164  0.849
Residuals       6374 228400   35.83               

The ANOVA test shows that there is no significant difference in the average number of hours students study per week based on teacher quality levels, with a p-value greater than 0.05. This indicates that teacher quality does not have a significant impact on the number of hours students study per week.

Average Number of Hours Students Study per Week (School Type Category)

Next, we explore the average number of hours students study per week based on school type.

# Summary statistics for study hours by school type
summary(student_data$Hours_Studied[student_data$School_Type == "Public"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   1.00   16.00   20.00   19.98   24.00   43.00 

summary(student_data$Hours_Studied[student_data$School_Type == "Private"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   1.00   16.00   20.00   19.97   24.00   44.00 

The summary statistics of the average number of hours students study per week based on school type show no significant differences. We will now use ANOVA to test for differences in study hours based on school type. But first, we need to check the assumptions of ANOVA.

# Sample the data
public_school <- student_data$Hours_Studied[student_data$School_Type == "Public"][sample(sum(student_data$School_Type == "Public"), 100)]
private_school <- student_data$Hours_Studied[student_data$School_Type == "Private"][sample(sum(student_data$School_Type == "Private"), 100)]

# Histogram of study hours by school type
par(mfrow = c(1, 2))
hist(public_school, main = "Public School", xlab = "Study Hours", col = "skyblue", border = "black")
hist(private_school, main = "Private School", xlab = "Study Hours", col = "skyblue", border = "black")

# Check assumptions of ANOVA
shapiro.test(public_school)

    Shapiro-Wilk normality test

data:  public_school
W = 0.98459, p-value = 0.2967

shapiro.test(private_school)

    Shapiro-Wilk normality test

data:  private_school
W = 0.98734, p-value = 0.4609

Both histograms and the Shapiro-Wilk test show that the distribution of study hours is approximately normal for all students. We will now use ANOVA to test for differences in study hours based on school type.

# ANOVA test for study hours by school type
anova_school_type <- aov(Hours_Studied ~ School_Type, data = student_data)
summary(anova_school_type)

              Df Sum Sq Mean Sq F value Pr(>F)
School_Type    1      0    0.25   0.007  0.934
Residuals   6375 228411   35.83               

The ANOVA test shows that there is no significant difference in the average number of hours students study per week based on school type, with a p-value greater than 0.05. This indicates that school type does not have a significant impact on the number of hours students study per week.

Average Number of Hours Students Study per Week (Peer Influence Category)

Next, we explore the average number of hours students study per week based on peer influence levels.

# Summary statistics for study hours by peer influence
summary(student_data$Hours_Studied[student_data$Peer_Influence == "Positive"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   1.00   16.00   20.00   20.06   24.00   43.00 

summary(student_data$Hours_Studied[student_data$Peer_Influence == "Negative"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   1.00   16.00   20.00   19.95   24.00   44.00 

summary(student_data$Hours_Studied[student_data$Peer_Influence == "Neutral"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   2.00   16.00   20.00   19.91   24.00   39.00 

The summary statistics of the average number of hours students study per week based on peer influence levels show no significant differences. We will now use ANOVA to test for differences in study hours based on peer influence levels. But first, we need to check the assumptions of ANOVA.

# Sample the data
positive_peer_influence <- student_data$Hours_Studied[student_data$Peer_Influence == "Positive"][sample(sum(student_data$Peer_Influence == "Positive"), 100)]
negative_peer_influence <- student_data$Hours_Studied[student_data$Peer_Influence == "Negative"][sample(sum(student_data$Peer_Influence == "Negative"), 100)]
neutral_peer_influence <- student_data$Hours_Studied[student_data$Peer_Influence == "Neutral"][sample(sum(student_data$Peer_Influence == "Neutral"), 100)]

# Histogram of study hours by peer influence
par(mfrow = c(1, 3))
hist(positive_peer_influence, main = "Positive Peer Influence", xlab = "Study Hours", col = "skyblue", border = "black")
hist(negative_peer_influence, main = "Negative Peer Influence", xlab = "Study Hours", col = "skyblue", border = "black")
hist(neutral_peer_influence, main = "Neutral Peer Influence", xlab = "Study Hours", col = "skyblue", border = "black")

# Check assumptions of ANOVA
shapiro.test(positive_peer_influence)

    Shapiro-Wilk normality test

data:  positive_peer_influence
W = 0.98724, p-value = 0.454

shapiro.test(negative_peer_influence)

    Shapiro-Wilk normality test

data:  negative_peer_influence
W = 0.98803, p-value = 0.5107

shapiro.test(neutral_peer_influence)

    Shapiro-Wilk normality test

data:  neutral_peer_influence
W = 0.97778, p-value = 0.08907

All three histograms and the Shapiro-Wilk test show that the distribution of study hours is approximately normal for all students. We will now use ANOVA to test for differences in study hours based on peer influence levels.

# ANOVA test for study hours by peer influence
anova_peer_influence <- aov(Hours_Studied ~ Peer_Influence, data = student_data)
summary(anova_peer_influence)

                 Df Sum Sq Mean Sq F value Pr(>F)
Peer_Influence    2     29   14.33     0.4   0.67
Residuals      6374 228383   35.83               

The ANOVA test shows that there is no significant difference in the average number of hours students study per week based on peer influence levels, with a p-value greater than 0.05. This indicates that peer influence does not have a significant impact on the number of hours students study per week.

Average Number of Hours Students Study per Week (Learning Disability Category)

Next, we explore the average number of hours students study per week based on learning disability.

# Summary statistics for study hours by learning disability
summary(student_data$Hours_Studied[student_data$Learning_Disabilities == "Yes"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   4.00   16.00   20.00   19.73   24.00   35.00 

summary(student_data$Hours_Studied[student_data$Learning_Disabilities == "No"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
      1      16      20      20      24      44 

The summary statistics of the average number of hours students study per week based on learning disability show no significant differences. We will now use ANOVA to test for differences in study hours based on learning disability. But first, we need to check the assumptions of ANOVA.

# Sample the data
learning_disabilities <- student_data$Hours_Studied[student_data$Learning_Disabilities == "Yes"][sample(sum(student_data$Learning_Disabilities == "Yes"), 100)]
no_learning_disabilities <- student_data$Hours_Studied[student_data$Learning_Disabilities == "No"][sample(sum(student_data$Learning_Disabilities == "No"), 100)]

# Histogram of study hours by learning disability
par(mfrow = c(1, 2))
hist(learning_disabilities, main = "Learning Disabilities", xlab = "Study Hours", col = "skyblue", border = "black")
hist(no_learning_disabilities, main = "No Learning Disabilities", xlab = "Study Hours", col = "skyblue", border = "black")

# Check assumptions of ANOVA
shapiro.test(learning_disabilities)

    Shapiro-Wilk normality test

data:  learning_disabilities
W = 0.98009, p-value = 0.135

shapiro.test(no_learning_disabilities)

    Shapiro-Wilk normality test

data:  no_learning_disabilities
W = 0.97305, p-value = 0.03802

All histograms and the Shapiro-Wilk test show that the distribution of study hours is approximately normal for all students. We will now use ANOVA to test for differences in study hours based on learning disability.

# ANOVA test for study hours by learning disability
anova_learning_disabilities <- aov(Hours_Studied ~ Learning_Disabilities, data = student_data)
summary(anova_learning_disabilities)

                        Df Sum Sq Mean Sq F value Pr(>F)
Learning_Disabilities    1     44   43.87   1.225  0.268
Residuals             6375 228367   35.82               

The ANOVA test shows that there is no significant difference in the average number of hours students study per week based on learning disability, with a p-value greater than 0.05. This indicates that learning disabilities do not have a significant impact on the number of hours students study per week.

Average Number of Hours Students Study per Week (Parental Education Level Category)

Next, we explore the average number of hours students study per week based on parental education levels.

# Summary statistics for study hours by parental education level
summary(student_data$Hours_Studied[student_data$Parental_Education_Level == "High School"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   1.00   16.00   20.00   20.05   24.00   44.00 

summary(student_data$Hours_Studied[student_data$Parental_Education_Level == "College"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   1.00   16.00   20.00   19.87   24.00   39.00 

summary(student_data$Hours_Studied[student_data$Parental_Education_Level == "Postgraduate"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   3.00   16.00   20.00   19.97   24.00   39.00 

The summary statistics of the average number of hours students study per week based on parental education levels show no significant differences. We will now use ANOVA to test for differences in study hours based on parental education levels. But first, we need to check the assumptions of ANOVA.

# Sample the data
high_school_education <- student_data$Hours_Studied[student_data$Parental_Education_Level == "High School"][sample(sum(student_data$Parental_Education_Level == "High School"), 100)]
college_education <- student_data$Hours_Studied[student_data$Parental_Education_Level == "College"][sample(sum(student_data$Parental_Education_Level == "College"), 100)]
postgraduate_education <- student_data$Hours_Studied[student_data$Parental_Education_Level == "Postgraduate"][sample(sum(student_data$Parental_Education_Level == "Postgraduate"), 100)]

# Histogram of study hours by parental education level
par(mfrow = c(1, 3))
hist(high_school_education, main = "High School Education", xlab = "Study Hours", col = "skyblue", border = "black")
hist(college_education, main = "College Education", xlab = "Study Hours", col = "skyblue", border = "black")
hist(postgraduate_education, main = "Postgraduate Education", xlab = "Study Hours", col = "skyblue", border = "black")

# Check assumptions of ANOVA
shapiro.test(high_school_education)

    Shapiro-Wilk normality test

data:  high_school_education
W = 0.98413, p-value = 0.2746

shapiro.test(college_education)

    Shapiro-Wilk normality test

data:  college_education
W = 0.97967, p-value = 0.1253

shapiro.test(postgraduate_education)

    Shapiro-Wilk normality test

data:  postgraduate_education
W = 0.98131, p-value = 0.168

All three histograms and the Shapiro-Wilk test show that the distribution of study hours is approximately normal for all students. We will now use ANOVA to test for differences in study hours based on parental education levels.

# ANOVA test for study hours by parental education level
anova_parental_education_level <- aov(Hours_Studied ~ Parental_Education_Level, data = student_data)
summary(anova_parental_education_level)

                           Df Sum Sq Mean Sq F value Pr(>F)
Parental_Education_Level    2     37   18.71   0.522  0.593
Residuals                6374 228374   35.83               

The ANOVA test shows that there is no significant difference in the average number of hours students study per week based on parental education levels, with a p-value greater than 0.05. This indicates that parental education levels do not have a significant impact on the number of hours students study per week.

Average Number of Hours Students Study per Week (Distance from Home Category)

Next, we explore the average number of hours students study per week based on the distance from home to school.

# Summary statistics for study hours by distance from home (Near, Moderate, Far
summary(student_data$Hours_Studied[student_data$Distance_from_Home == "Near"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
    1.0    16.0    20.0    19.9    24.0    43.0 

summary(student_data$Hours_Studied[student_data$Distance_from_Home == "Moderate"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   2.00   16.00   20.00   20.05   24.00   44.00 

summary(student_data$Hours_Studied[student_data$Distance_from_Home == "Far"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
    3.0    16.0    20.0    20.2    24.0    39.0 

The summary statistics of the average number of hours students study per week based on the distance from home to school show no significant differences. We will now use ANOVA to test for differences in study hours based on the distance from home to school. But first, we need to check the assumptions of ANOVA.

# Sample the data
near_distance <- student_data$Hours_Studied[student_data$Distance_from_Home == "Near"][sample(sum(student_data$Distance_from_Home == "Near"), 100)]
moderate_distance <- student_data$Hours_Studied[student_data$Distance_from_Home == "Moderate"][sample(sum(student_data$Distance_from_Home == "Moderate"), 100)]
far_distance <- student_data$Hours_Studied[student_data$Distance_from_Home == "Far"][sample(sum(student_data$Distance_from_Home == "Far"), 100)]

# Histogram of study hours by distance from home
par(mfrow = c(1, 3))
hist(near_distance, main = "Near Distance from Home", xlab = "Study Hours", col = "skyblue", border = "black")
hist(moderate_distance, main = "Moderate Distance from Home", xlab = "Study Hours", col = "skyblue", border = "black")
hist(far_distance, main = "Far Distance from Home", xlab = "Study Hours", col = "skyblue", border = "black")

# Check assumptions of ANOVA
shapiro.test(near_distance)

    Shapiro-Wilk normality test

data:  near_distance
W = 0.97892, p-value = 0.1094

shapiro.test(moderate_distance)

    Shapiro-Wilk normality test

data:  moderate_distance
W = 0.99076, p-value = 0.7258

shapiro.test(far_distance)

    Shapiro-Wilk normality test

data:  far_distance
W = 0.98969, p-value = 0.6399

All three histograms and the Shapiro-Wilk test show that the distribution of study hours is approximately normal for all students. We will now use ANOVA to test for differences in study hours based on the distance from home to school.

# ANOVA test for study hours by distance from home
anova_distance_from_home <- aov(Hours_Studied ~ Distance_from_Home, data = student_data)
summary(anova_distance_from_home)

                     Df Sum Sq Mean Sq F value Pr(>F)
Distance_from_Home    2     67   33.48   0.934  0.393
Residuals          6374 228344   35.82               

The ANOVA test shows that there is no significant difference in the average number of hours students study per week based on the distance from home to school, with a p-value greater than 0.05. This indicates that the distance from home to school does not have a significant impact on the number of hours students study per week.

Average Number of Hours Students Study per Week (Gender Category)

Finally, we explore the average number of hours students study per week based on gender.

# Summary statistics
summary(student_data$Hours_Studied[student_data$Gender == "Male"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   1.00   16.00   20.00   19.94   24.00   39.00 

summary(student_data$Hours_Studied[student_data$Gender == "Female"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   1.00   16.00   20.00   20.02   24.00   44.00 

The summary statistics of the average number of hours students study per week based on gender show no significant differences. We will now use ANOVA to test for differences in study hours based on gender. But first, we need to check the assumptions of ANOVA.

# Sample the data
male_data <- student_data$Hours_Studied[student_data$Gender == "Male"][sample(sum(student_data$Gender == "Male"), 100)]
female_data <- student_data$Hours_Studied[student_data$Gender == "Female"][sample(sum(student_data$Gender == "Female"), 100)]

# Histogram
par(mfrow = c(1, 2))
hist(male_data, main = "Male", xlab = "Study Hours", col = "skyblue", border = "black")
hist(female_data, main = "Female", xlab = "Study Hours", col = "skyblue", border = "black")

# Check assumptions of ANOVA
shapiro.test(male_data)

    Shapiro-Wilk normality test

data:  male_data
W = 0.99016, p-value = 0.6773

shapiro.test(female_data)

    Shapiro-Wilk normality test

data:  female_data
W = 0.9903, p-value = 0.6889

Both histograms and the Shapiro-Wilk test show that the distribution of study hours is approximately normal for all students. We will now use ANOVA to test for differences in study hours based gender.

# ANOVA test
anova_gender <- aov(Hours_Studied ~ Gender, data = student_data)
summary(anova_gender)

              Df Sum Sq Mean Sq F value Pr(>F)
Gender         1     11   11.12    0.31  0.577
Residuals   6375 228400   35.83               

The ANOVA test shows that there is no significant difference in the average number of hours students study based on gender.

Summary of Average Number of Hours Students Study per Week

In summary, the average number of hours students study per week is approximately 19.98 hours. There are no significant differences in the average number of hours students study per week based on various factors such as parental involvement, access to resources, extracurricular activities, motivation levels, internet access, family income, teacher quality, school type, peer influence, learning disabilities, parental education levels, distance from home to school, and Gender. This indicates that the number of hours students study per week is consistent across different factors.

The variation in attendance rates across students

Now we analyze the variation in attendance rates across students.

# Sample the data
sample_attendance <- student_data$Attendance[sample(nrow(student_data), 1000)]

# Summary statistics for attendance rates
summary(sample_attendance)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   60.0    70.0    81.0    80.8    92.0   100.0 

# Histogram of attendance rates
hist(sample_attendance, main = "Distribution of Attendance Rates", xlab = "Attendance Rate", col = "skyblue", border = "black")

From the histogram we can see that the distribution of attendance rates is approximately uniform, ranging from approximately 60% to 100%. We can see one slightly higher peak around 70% to 80%. From this we can see that attendance rates are evenly distributed across students. The average attendance rate is approximately 79.75%, which is relatively high. We now investigate the factors that may influence attendance rates.

Attendance Rates by Parental Involvement

We will now explore the attendance rates based on parental involvement levels.

# Summary statistics for attendance rates by parental involvement
summary(student_data$Attendance[student_data$Parental_Involvement == "High"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  60.00   70.00   80.00   80.01   90.00  100.00 

summary(student_data$Attendance[student_data$Parental_Involvement == "Medium"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   60.0    70.0    80.0    79.9    90.0   100.0 

summary(student_data$Attendance[student_data$Parental_Involvement == "Low"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  60.00   70.00   80.00   80.32   91.00  100.00 

The summary statistics of the attendance rates based on parental involvement levels show no significant differences. We will now use ANOVA to test for differences in attendance rates based on parental involvement levels. But first, we need to check the assumptions of ANOVA.

# Sample the data
high_parental_involvement_attendance <- student_data$Attendance[student_data$Parental_Involvement == "High"][sample(sum(student_data$Parental_Involvement == "High"), 100)]
medium_parental_involvement_attendance <- student_data$Attendance[student_data$Parental_Involvement == "Medium"][sample(sum(student_data$Parental_Involvement == "Medium"), 100)]
low_parental_involvement_attendance <- student_data$Attendance[student_data$Parental_Involvement == "Low"][sample(sum(student_data$Parental_Involvement == "Low"), 100)]

# Histogram of attendance rates by parental involvement
par(mfrow = c(1, 3))
hist(high_parental_involvement_attendance, main = "High Parental Involvement", xlab = "Attendance Rate", col = "skyblue", border = "black")
hist(medium_parental_involvement_attendance, main = "Medium Parental Involvement", xlab = "Attendance Rate", col = "skyblue", border = "black")
hist(low_parental_involvement_attendance, main = "Low Parental Involvement", xlab = "Attendance Rate", col = "skyblue", border = "black")

# Check assumptions of ANOVA
shapiro.test(high_parental_involvement_attendance)

    Shapiro-Wilk normality test

data:  high_parental_involvement_attendance
W = 0.9466, p-value = 0.0004986

shapiro.test(medium_parental_involvement_attendance)

    Shapiro-Wilk normality test

data:  medium_parental_involvement_attendance
W = 0.92294, p-value = 2.023e-05

shapiro.test(low_parental_involvement_attendance)

    Shapiro-Wilk normality test

data:  low_parental_involvement_attendance
W = 0.928, p-value = 3.842e-05

All three histograms and the Shapiro-Wilk test show that the distribution of attendance rates is not normal for all students. We will now use Kruksal-Wallis test to test for differences in attendance rates based on parental involvement levels.

# Kruksal-Wallis test
kruskal.test(Attendance ~ Parental_Involvement, data = student_data)

    Kruskal-Wallis rank sum test

data:  Attendance by Parental_Involvement
Kruskal-Wallis chi-squared = 1.2171, df = 2, p-value = 0.5441

The Kruskal-Wallis test shows that there is no significant difference in the average attendance rates based on parental involvement levels, with a p-value greater than 0.05. This indicates that parental involvement does not have a significant impact on attendance rates.

Attendance Rates by Access to Resources

Next, we explore the attendance rates based on access to resources.

# Summary statistics for attendance rates by access to resources
summary(student_data$Attendance[student_data$Access_to_Resources == "High"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  60.00   70.00   79.00   79.87   90.00  100.00 

summary(student_data$Attendance[student_data$Access_to_Resources == "Medium"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
     60      70      80      80      90     100 

summary(student_data$Attendance[student_data$Access_to_Resources == "Low"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  60.00   71.00   80.00   80.29   91.00  100.00 

The summary statistics of the attendance rates based on access to resources show no significant differences. We will now use ANOVA to test for differences in attendance rates based on access to resources. But first, we need to check the assumptions of ANOVA.

# Sample the data
high_access_to_resources_attendance <- student_data$Attendance[student_data$Access_to_Resources == "High"][sample(sum(student_data$Access_to_Resources == "High"), 100)]
medium_access_to_resources_attendance <- student_data$Attendance[student_data$Access_to_Resources == "Medium"][sample(sum(student_data$Access_to_Resources == "Medium"), 100)]
low_access_to_resources_attendance <- student_data$Attendance[student_data$Access_to_Resources == "Low"][sample(sum(student_data$Access_to_Resources == "Low"), 100)]

# Histogram of attendance rates by access to resources
par(mfrow = c(1, 3))
hist(high_access_to_resources_attendance, main = "High Access to Resources", xlab = "Attendance Rate", col = "skyblue", border = "black")
hist(medium_access_to_resources_attendance, main = "Medium Access to Resources", xlab = "Attendance Rate", col = "skyblue", border = "black")
hist(low_access_to_resources_attendance, main = "Low Access to Resources", xlab = "Attendance Rate", col = "skyblue", border = "black")

# Check assumptions of ANOVA
shapiro.test(high_access_to_resources_attendance)

    Shapiro-Wilk normality test

data:  high_access_to_resources_attendance
W = 0.95112, p-value = 0.0009836

shapiro.test(medium_access_to_resources_attendance)

    Shapiro-Wilk normality test

data:  medium_access_to_resources_attendance
W = 0.94251, p-value = 0.0002752

shapiro.test(low_access_to_resources_attendance)

    Shapiro-Wilk normality test

data:  low_access_to_resources_attendance
W = 0.93998, p-value = 0.0001923

All three histograms and the Shapiro-Wilk test show that the distribution of attendance rates is not normal for all students. We will now use Kruksal-Wallis test to test for differences in attendance rates based on access to resources.

# Kruksal-Wallis test
kruskal.test(Attendance ~ Access_to_Resources, data = student_data)

    Kruskal-Wallis rank sum test

data:  Attendance by Access_to_Resources
Kruskal-Wallis chi-squared = 1.0494, df = 2, p-value = 0.5917

The Kruskal-Wallis test shows that there is no significant difference in the average attendance rates based on access to resources, with a p-value greater than 0.05. This indicates that access to resources does not have a significant impact on attendance rates.

Attendance Rates by Extracurricular Activities

Next, we explore the attendance rates based on participation in extracurricular activities.

# Summary statistics for attendance rates by extracurricular activities
summary(student_data$Attendance[student_data$Extracurricular_Activities == "Yes"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
     60      70      80      80      90     100 

summary(student_data$Attendance[student_data$Extracurricular_Activities == "No"])

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  60.00   70.00   80.00   80.05   90.00  100.00 

The summary statistics of the attendance rates based on participation in extracurricular activities show no significant differences. We will now use ANOVA to test for differences in attendance rates based on participation in extracurricular activities. But first, we need to check the assumptions of ANOVA.

# Sample the data
participate_extracurricular_attendance <- student_data$Attendance[student_data$Extracurricular_Activities == "Yes"][sample(sum(student_data$Extracurricular_Activities == "Yes"), 100)]
do_not_participate_extracurricular_attendance <- student_data$Attendance[student_data$Extracurricular_Activities == "No"][sample(sum(student_data$Extracurricular_Activities == "No"), 100)]

# Histogram of attendance rates by extracurricular activities
par(mfrow = c(1, 2))
hist(participate_extracurricular_attendance, main = "Extracurricular Activities", xlab = "Attendance Rate", col = "skyblue", border = "black")
hist(do_not_participate_extracurricular_attendance, main = "No Extracurricular Activities", xlab = "Attendance Rate", col = "skyblue", border = "black")

# Check assumptions of ANOVA
shapiro.test(participate_extracurricular_attendance)

    Shapiro-Wilk normality test

data:  participate_extracurricular_attendance
W = 0.96044, p-value = 0.004321

shapiro.test(do_not_participate_extracurricular_attendance)

    Shapiro-Wilk normality test

data:  do_not_participate_extracurricular_attendance
W = 0.93609, p-value = 0.0001122

Both histograms and the Shapiro-Wilk test show that the distribution of attendance rates is not normal for all students. We will now use Kruksal-Wallis test to test for differences in attendance rates based on participation in extracurricular activities.

# Kruksal-Wallis test
kruskal.test(Attendance ~ Extracurricular_Activities, data = student_data)

    Kruskal-Wallis rank sum test

data:  Attendance by Extracurricular_Activities
Kruskal-Wallis chi-squared = 0.026414, df = 1, p-value = 0.8709

The Kruskal-Wallis test shows that there is no significant difference in the average attendance rates based on participation in extracurricular activities, with a p-value greater than 0.05. This indicates that participation in extracurricular activities does not have a significant impact on attendance rates.